我之前问过类似的问题,但我需要一些进一步的输出,并决定发布一个新问题。
我有一个像这样的 data.table 对象:
library(data.table)
cells <- c(100, 1,1980,1,0,1,1,0,1,0,
150, 1,1980,1,1,1,0,0,0,1,
99 , 1,1980,1,1,1,1,0,0,0,
899, 1,1980,0,1,0,1,1,1,1,
789, 1,1982,1,1,1,0,1,1,1 )
colname <- c("number","sex", "birthy", "2004","2005", "2006", "2007", "2008", "2009","2010")
rowname <- c("1","2","3","4","5")
y <- matrix(cells, nrow=5, ncol=10, byrow=TRUE, dimnames = list(rowname,colname))
y <- data.table(y, keep.rownames = TRUE)
2004 列中的值 1 表示此人在 2004 年期间连续受保。之前 3 年受保的人可以参与研究。我需要此 data.table 的一个子集,其中包含满足以下条件的所有观察结果:2004+2005+2006 = 3 或 2005+2006+2007 = 或 2006+2007+...
#using melt and rle function to restrucure the data
tmp <- melt(y, id = "rn", measure.vars = patterns("^20"),
variable.factor = FALSE, variable.name = "year")[, rle(value), by = rn]
#subset data based on condition, keeping only the first relevant sequence
tmp2 <- tmp[(values == 1 & lengths >= 3), .(rn,lengths)][, .SD[1,], by=rn]
##selecting only rows with value=1 and min 3 in a row
##keeping only the variable rn
tmp3 <- tmp[values == 1, which(max(lengths) >= 3), by = rn]$rn
##using the row-number to select obersvations from data.table
##merging length of sequence
dt <- merge(y[as.integer(tmp3)],tmp2, by="rn")
如果它们不是序列的一部分,有没有办法将所有 1 变为 0?例如 rn==4 变量“2005”需要为零。
我还需要一个新变量“begy”,其中包含序列开始的年份。例如rn==5
和begy==2004
。任何建议将不胜感激...
最佳答案
新解决方案:
# define a custom function in order to only keep the sequences
# with 3 (or more) consecutive years
rle3 <- function(x) {
r <- rle(x)
r$values[r$lengths < 3 & r$values == 1] <- 0
inverse.rle(r)
}
# replace all '1'-s that do not belong to a sequence of at least 3 to '0'
# create 'begy'-variable
melt(y, id = 1:4, measure.vars = patterns("^20"),
variable.factor = FALSE, variable.name = "year"
)[, value := rle3(value), by = rn
][, begy := year[value == 1][1], rn
][, dcast(.SD[!is.na(begy)], ... ~ year, value.var = "value")]
给出:
rn number sex birthy begy 2004 2005 2006 2007 2008 2009 2010 1: 2 150 1 1980 2004 1 1 1 0 0 0 0 2: 3 99 1 1980 2004 1 1 1 1 0 0 0 3: 4 899 1 1980 2007 0 0 0 1 1 1 1 4: 5 789 1 1982 2004 1 1 1 0 1 1 1
旧解决方案:
# define a custom function in order to only keep the sequences
# with 3 (or more) consecutive years
rle3 <- function(x) {
r <- rle(x)
r$values[r$lengths < 3 & r$values == 1] <- 0
inverse.rle(r)
}
# create a reference 'data.table' with only the row to keep
# and the start year of the (first) sequence (row 5 has 2 sequences of 3)
x <- melt(y, id = "rn", measure.vars = patterns("^20"),
variable.factor = FALSE, variable.name = "year"
)[, value := rle3(value), by = rn
][value == 1, .SD[1], rn]
# join 'x' with 'y' to add 'begy' and filter out the row with no sequences of 3
y[x, on = "rn", begy := year][!is.na(begy)]
给出:
rn number sex birthy 2004 2005 2006 2007 2008 2009 2010 begy 1: 2 150 1 1980 1 1 1 0 0 0 1 2004 2: 3 99 1 1980 1 1 1 1 0 0 0 2004 3: 4 899 1 1980 0 1 0 1 1 1 1 2007 4: 5 789 1 1982 1 1 1 0 1 1 1 2004
关于r - 如果 Sequence <= 3 则将所有值解码为零,并保留某些信息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52034420/